# How do you solve log_5(3x-1)=log_5 (2x^2) and check the solutions?

Dec 18, 2016

$x = \frac{1}{2} \mathmr{and} 1$

#### Explanation:

We start by using the property that if color(magenta)(log_a b = log_a c, then $\textcolor{m a \ge n t a}{b = c}$.

Therefore:

$3 x - 1 = 2 {x}^{2}$

$0 = 2 {x}^{2} - 3 x + 1$

$0 = 2 {x}^{2} - 2 x - x + 1$

$0 = 2 x \left(x - 1\right) - 1 \left(x - 1\right)$

$0 = \left(2 x - 1\right) \left(x - 1\right)$

$x = \frac{1}{2} \mathmr{and} 1$

Checking in the original equation, you will find both solutions work. Note that our restrictions on the variable are $x > 0$, because the log function is undefined in the real number system whenever the value within the logarithm is equal to or smaller than $0$. We can automatically confirm both solutions are correct because neither contradict the restriction.

Hopefully this helps!